ESE 570 MOS TRANSISTOR THEORY Part 1. Kenneth R. Laker, University of Pennsylvania, updated 5Feb15

Transcription

1 ESE 570 MOS TRANSISTOR THEORY Part 1

2 TwoTerminal MOS Structure 2 GATE Si Oxide interface n n Mass Action Law VB 2

3 Chemical Periodic Table Donors American Chemical Society (ACS) Acceptors Metalloids 3

4 Ideal Equilibrium MOS Capacitor Energy Bands q / Si =q 3 Si (,E C E V Oxide qφm metal VG = 0 Work Functions qφm, qφsi = energy required to move an electron from E F to Evacuum for metal gate, Si respectively. Si surface EC, EFm EFp Gate p doped Si E C E V E i= 2 NOTE: 1. qφ and E are in units of energy = electronvolts (ev); where 1 ev = 1.6 x J ev corresponds to energy acquired by a free electron that is accelerated by an electric potential of one volt. 3. Φ and V corresponds to potential difference in volts. 4

5 MOS Capacitor with External Bias Three Regions of Operation: 1. Accumulation Region VG < 0 2. Depletion Region VG > 0, small 3. Inversion Region VG VT, large 5

6 Energy Bands Accumulation Region Si surface Accumulation VG < 0 EFm qv G =E Fp E Fm Band bending due to VG < 0 qφ(x) qφs qφfp 0 EFp x q / Fp = E Fp E i bulk )0 q /, x = E Fp E i, x Surface potential: q / S =q /,0)0 Band bending:. E i,x = E i,x E i bulk *0 6

7 MOS Capacitor Depletion Region tox mobile holes 7

8 Energy Bands Depletion Region Si surface Depletion VG > 0 (small) Band bending due to VG > 0 qφ(x) qφfp qv G =E Fp E Fm qφs EFm xd EFp q / Fp = E Fp E i bulk )0 q /, x = E Fp E i, x x 0 Surface Potential: q / S =q /,0*0 Band bending:. E i,x = E i, x E i bulk )0 8

9 MOS Capacitor Depletion Region tox surface potential (Fermi 2 S potential at surface) 2 2 Fp Bulk or Fermi potential 2 =2 = kt ln ni )0 Fp F q NA 26 mv at room T Mobile hole charge density (per unit area) in thin layer parallel to SiOxide interface Depletion region potential needed to displace dq by distance x into bulk (Poisson Eq.) NOTE 2 Fp 2Fp 2S 2S 2 FS 2 1Si 2 2FpS 2 xd= qna Q= q N A x d = 2 q N A 1 Si 2 2S 2 F S Fp 2 9

10 Energy Bands Inversion Region Si surface Inversion VG VT0 > 0 / S = / V G =V T0 qv G =E Fp E Fm qφfp qφs EFm EFp q / Fp = E Fp E i bulk )0 xdm q /, x = E Fp E i, x x 0 Surface Potential: q/ S =q /,0= q / Fp *0 Band bending:. E i,x = E i, x E i bulk )0 10

11 MOS Capacitor Inversion Region VG VT (threshold voltage) tox VG = VT for φs = φfp F.= q N A 1 Si Fp S = 2 Fp 22S = 2 F 2 1 Si 2 2Fp 2 1 Si 2 2 F2 S 2F x dm= x dm=x d l 2 = 22S= q N A = 2F qn A ni kt 2 Fp =2 F = ln q NA S F,2 S = 2 F kt N D 2Fn =2 F = ln q ni 11

12 MOS Capacitor Inversion Region VG VT (threshold voltage) INVERSION CONDITION Key Equations 2S = 2F when n = NA ni kt 2 F =2 Fp= ln V q NA kt N D V 2F =2 Fn = ln q ni Depletion region charge density psub nsub Q B0 = qn A 1 Si 2 2 F Where c/cm2 1 Si 31 ox F/cm VG = VFB for φs = φf flat band (FB) condition, i.e. no band bending. 12

13 nmos layout 13

14 G 14

15 TwoTerminal MOS Structure > nmos Transistor VG VS VD depletion region 15

16 nmos Transistor = MOS Capacitor source/drain VSB = 0 VS VD VG NOTE: In Cadence SPICE = Spectre 1 Si 31 ox m where 21Si 2 2Fp V SB where x dm= qn A NOTE: Since NA >> ni : φfp < 0 16

17 VT0n,p [VT0 > VT0 in SPICE] Q ox Q B0 for nmos and pmos V T0=/GC 22 F VFB = flat band voltage Q ox V FB =/GC / GC Q B0 = q N A 1 Si 2 2 F ) with VSB = 0. VFB VFB l work function between gate and channel for psub 2 F =2 Fp 17

18 Adjusting VT0 Using and an Added Channel Implant Q B0 V T0 =V FB 2 2F Intrinsic VT0 no channel implant adjustment Q B0 q N I V =V T0 (. V T0 =V FB 22 F ± ' T0 qni.v T0 =± Adjusted V'T0 due to channel implant adjustment with carrier concentration NI qni for ptype implant ( q N I for ntype implant NOTE: When channel implant adjustment N I is done as a step in the CMOS process, the SPICE parameter VT0 refers to the adjusted threshold voltage V'T0. 18

19 Q B0 V T0 =V FB 2 2F Q B= q N A 1 Si 2 2 F V SB for VSB = 0 Q B0 = q N A 1 Si 2 2F Q B Q B0 Q B0 ( V T =V FB 22 F Q B0 Q B Q B0 V T =V FB 22 F VT0 Q B Q B0 2q N A 1 Si =, 2 2F V SB 2 2 F V T =V T0 (0, 2 2 F V SB 2 2 F units = V1/2 19

20 VSB is 0 in nmos, 0 in pmos VOX is negative positive in )pmos, negative in pmos (VT0p) Q is fornmos nmos(v and T0n T0 0 20

21 VSB 21

22 1 for 1 A=10 10 m 1 φf Q B0 V T0n=V FB 22Fp ni kt 1.45 x Fp = ln =0.26V ln, = 0.35V 16 q NA 10 22

23 Q B0 V T0n=V FB 22Fp 1 2 Fp = 0.35 V Q B0 = 2 q N A 1 Si 22 Fp.= 2,1.6 x C,10 cm,1.06 x 10 1 F cm 0.70 V F = C/V V T0n= 1.04 V 2, 0.35V, 0.72 V =0.38 V 23

24 2 Example 1 2Fp = 0.35V bulk potential V Tn=V T0n(0, 2 2 Fp V SB 2 2 F Units Calc x 10 8 C /,V 1/ 2 cm 2 1/ 2.= =0.85 V 1/2 V 6.8 x 10 8 C /,V cm 2 C 2 cm 4 V 1 =V 1/ 2 2 C cm V 1 24

25 2 V Tn=V T0n(0, 22Fp V SB 22 F where V T0n =0.38 V 1/ 2 0=0.85V 2Fp = 0.35V 1 V Tn=0.38 V (0, 0.70 V V SB 0.70 V 25